Modern Physics 1 Question 43
40. The recoil speed of a hydrogen atom after it emits a photon is going from $n=5$ state to $n=1$ state is $m / s . \quad(1997,1 M)$
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Solution:
- From conservation of linear momentum,
|Momentum of recoil hydrogen atom $|=|$ Momentum of emitted photon|
or $\quad m v=\frac{\Delta E}{c}$
Here, $\Delta E=E _5-E _1=-13.6 \frac{1}{5^{2}}-\frac{1}{1^{2}} eV$
$$ \begin{aligned} & =(13.6)(24 / 25) eV=13.056 eV \\ & =13.056 \times 1.6 \times 10^{-19} J=2.09 \times 10^{-18} J \end{aligned} $$
and $m=$ mass of hydrogen atom $=1.67 \times 10^{-27} kg$
$$ \begin{aligned} \therefore \quad v & =\frac{\Delta E}{m c}=\frac{2.09 \times 10^{-18}}{\left(1.67 \times 10^{-27}\right)\left(3 \times 10^{8}\right)}, \\ v & \approx 4.17 m / s \end{aligned} $$
